# Stochastic Calculus

## Sum of Centered Gaussians

Consider the sum of two independent random variables $X$ and $Y$, where each variable is normally distributed with zero mean. Denote the variance of $X$ as $\sigma^{2}_{X}$ and the variance of $Y$ as $\sigma^{2}_{Y}$. $X \sim \mathcal{N}(0, \sigma^{2}_{X}) \quad ; \quad Y \sim \mathcal{N}(0, \sigma^{2}_{Y}).$ Recall that the sum of $X$ and $Y$ is distributed with variance $\sigma^{2}_{X} + \sigma^{2}_{Y}$.

$$$\label{eqn-1}\tag{1} Z = X + Y \implies Z \sim \mathcal{N}(0, \sigma^{2}_{X} + \sigma^{2}_{Y}).$$$

## A Continuous Random Walk

Imagine a continuous-time random walk $W(t)$ for which observed differences between $W(t_{i})$ and $W(t_j)$ are always normally distributed with zero mean:

$$$\label{eqn-2}\tag{2} W(t_j) - W(t_i) \sim \mathcal{N}(0, \sigma^{2}_{ij}).$$$

Assume also that the random walk is uncorrelated with itself when comparing non-overlapping time intervals.

Consistency between Eq. (2) and Eq. (1), i.e., $\forall ~ t_{i} < t_{j} < t_{k}, \quad \sigma^{2}_{ij} + \sigma^{2}_{jk} = \sigma^{2}_{ik},$ dictates that $\sigma^{2}_{ij} \propto |t_j - t_i|.$ Let us choose to define $W$ such that the constant of proportionality is 1, identifying this random walk as a Wiener process. $W(t_{j}) - W(t_{i}) \sim \mathcal{N}\big(0, |t_{j} - t_{i}|\big).$ For positive, infinitesimal differences in time, we may write

$$$\label{eqn-3}\tag{3} {\rm d}W \sim \mathcal{N}(0, {\rm d} t).$$$

Recall that, for any random variable $Z$, ${\rm Var}[Z] \equiv \mathbb{E}[Z^{2}] - \mathbb{E}[Z]^{2}.$ For a Wiener process, this implies ${\rm Var}[{\rm d}W] = \mathbb{E}[({\rm d}W)^{2}].$ We may therefore rewrite Eq. (3) in terms of a quadratic variation: $\mathbb{E}[({\rm d}W)^{2}] = {\rm d}t.$ This observation has an important consequence: Approximations of a process $X(t)$ depending on $W(t)$, to first-order in ${\rm d}t$, must account for the quadratic variation in $W$. That is,

\label{eqn-4}\tag{4} \begin{aligned} {\rm d}X(t) &\approx {\frac{{\partial} X}{{\partial} t}}{\rm d}t + {\frac{{\partial} X}{{\partial} W}}{\rm d}W(t) + \frac{1}{2} {\frac{{\partial}^{2} X}{{\partial} W^{2}}} ({\rm d}W(t))^{2} \\ &= \bigg( {\frac{{\partial} X}{{\partial} t}} + \underbrace{\frac{1}{2} {\frac{{\partial}^{2} X}{{\partial} W^{2}}}}_{\mkern-1.5em\text{quadratic dependence!}\mkern-1.5em} \bigg) {\rm d}t + {\frac{{\partial} X}{{\partial} W}}{\rm d}W(t). \end{aligned}

## Drift-Diffusion Processes

An Itô drift-diffusion process may be represented in terms of differentials in $t$ and $W$ as ${\rm d}X(t) = \mu_{X}(t){\rm d}t + \sigma_{X}(t) {\rm d}W(t),$ where $\mu_{X}(t)$ and $\sigma_{X}(t)$ are given by

\begin{align} \mu_{X}(t) &= {\frac{{\partial} X}{{\partial} t}} + \frac{1}{2} {\frac{{\partial}^{2} X}{{\partial} W^{2}}}. \\ \sigma_{X}(t) &= {\frac{{\partial} X}{{\partial} W}}. \end{align}

The functions $\mu_{X}$ and $\sigma_{X}$ are deterministic (e.g., as functions of $t$ and the history of $X$), while $W(t)$ is stochastic (a Wiener process as described above). Regarding notation, observe that $\mu_{X}(t)$ and $\sigma^{2}(t)$ are the mean and variance, respectively, for ${\rm d}X$, not for $X$!

### Itô’s Lemma

Twice-differentiable functions applied to drift-diffusion stochastic processes also define drift-diffusion stochastic processes. For example, consider a twice-differentiable function $f(t, x)$ where the second argument is given by a stochastic process $X$, such that ${\rm d}X(t) = \mu_{X}(t){\rm d}t + \sigma_{X}(t) {\rm d}W(t).$ We may express $F = f(t, X)$ as a drift-diffusion process: ${\rm d}F(t) = \mu_{F}(t){\rm d}t + \sigma_{F}(t) {\rm d}W(t).$ To relate the factors of $\mu_{F}, \mu_{X}, \sigma_{F}$, and $\sigma_{X}$, first Taylor expand $F$ according to Eq. (4), i.e., ${\rm d}F = \bigg( \Big({\frac{{\partial} F}{{\partial} t}}\Big)_{W} + \frac{1}{2} {\frac{{\partial}^{2} F}{{\partial} W^{2}}} \bigg) {\rm d}t + {\frac{{\partial} F}{{\partial} W}}{\rm d}W(t).$ Next, apply the chain rule to second-order, i.e.,

\begin{align} \frac{{\partial}F}{{\partial}W} &= \frac{{\partial}f}{{\partial}x} \frac{{\partial}X}{{\partial}W}; \quad \Big(\frac{{\partial}F}{{\partial}t}\Big)_{W} = \frac{\partial f}{\partial t} + \frac{{\partial}f}{{\partial}x} \frac{{\partial}X}{{\partial}t}; \\ \frac{{\partial}^{2} F}{{\partial} W^{2}} &= \frac{{\partial}^{2} f}{{\partial} x^{2}} \bigg(\frac{{\partial} X}{{\partial} W}\bigg)^{2} + \frac{{\partial} f}{{\partial} x} \frac{{\partial}^{2} X}{{\partial} W^{2}}. \end{align}

We see that ${\rm d}F = \bigg( \frac{{\partial}f}{{\partial}t} + \frac{{\partial}f}{{\partial}x} \bigg( \underbrace{ \frac{{\partial}X}{{\partial}t} + \frac{1}{2} \frac{{\partial}^{2} X}{{\partial} W^{2}}}_{\mu_{X}} \bigg) + \frac{1}{2} \frac{{\partial}^{2} f}{{\partial} x^{2}} \bigg(\underbrace{\frac{{\partial} X}{{\partial} W}}_{\sigma_{X}}\bigg)^{2} \bigg) {\rm d}t + \frac{{\partial}f}{{\partial}x} \underbrace{\frac{{\partial}X}{{\partial}W}}_{\sigma_{X}} {\rm d}W(t).$ We have thus derived Itô’s Lemma: $\mu_{f}(t) = {\frac{\partial f}{\partial t}} + {\frac{\partial f}{\partial x}} \mu_{X}(t) + {\frac{1}{2} \frac{\partial^{2} f}{\partial x^{2}}} \sigma^2_{X}(t) \quad ; \quad \sigma_{f}(t) = {\frac{\partial f}{\partial x}} \sigma_{X}(t).$ Importantly, our result differs from the classical chain rule! ${\rm d}F = \frac{\partial f}{\partial t} {\rm d}t + \frac{\partial f}{\partial x} {\rm d}X + \underbrace{ \frac{1}{2} \frac{\partial^{2} f}{\partial x^{2}} \bigg(\frac{\partial X}{\partial W} \bigg)^{2} }_{\text{non-classical term}}.$

## Geometric Brownian Motion

Consider a stochastic process $X$ for which the proportional growth $\frac{{\rm d}X}{X}$ is an affine transformation of a Wiener process, i.e.,

$$$\label{eqn-7}\tag{7} {\rm d}X(t) = X(t) \bigg( \mu{\rm d}t + \sigma {\rm d}W(t) \bigg)$$$

We provide two means of solving for $X(t)$:

### By Itô’s Lemma

We have the drift-diffusion process ${\rm d}X = (\mu X) {\rm d}t + (\sigma X) {\rm d}W.$ Let us apply Itô’s Lemma for the mapping $f(t, x) = \log x$, where $F = f(t, X)$: ${\rm d}F = \bigg( \frac{\partial f}{\partial x} (\mu X) + \frac{1}{2}\frac{\partial^{2} f}{\partial x^{2}} (\sigma X)^{2} \bigg) {\rm d}t + \frac{\partial f}{ \partial x} (\sigma X) {\rm d}W.$ Substituting $\frac{\partial f}{\partial x}\bigg\rvert_{x{=}X} = \frac{1}{X} \quad ; \quad \frac{\partial^{2} f}{\partial x^{2}}\bigg\rvert_{x{=}X} = -\frac{1}{X^{2}},$ we obtain ${\rm d}F = \bigg( \mu - \frac{1}{2} \sigma^{2} \bigg) {\rm d}t + \sigma {\rm d}W.$ For constant $\mu, \sigma$, this differential equation has solution $F(t) = \bigg(\mu - \frac{1}{2}\sigma^{2}\bigg) t + \sigma W(t) + C,$ where $C$ is given by boundary conditions (i.e., the value of $X(0)$). Substituting $X(t) = X_{0} e^{F(t)}$, we conclude $X(t) = X_{0} e^{\big(\mu - \frac{1}{2}\sigma^{2}\big) t + \sigma W(t)}.$

Another approach to solving for $X$ is to note that both the definition of geometric Brownian motion (Eq. (7)) and a Taylor expansion of $X$ (Eq. (4)) must be consistent.

\begin{align*} {\rm d}X(t) &= X(t) \bigg( \mu{\rm d}t + \sigma {\rm d}W(t) \bigg). \\ {\rm d}X(t) &= \bigg( {\frac{{\partial} X}{{\partial} t}} + \frac{1}{2} {\frac{{\partial}^{2} X}{{\partial} W^{2}}} \bigg) {\rm d}t + {\frac{{\partial} X}{{\partial} W}}{\rm d}W(t). \end{align*}

By the independence of ${\rm d}t$ and ${\rm d}W$, this provides two equations: $\mu X = {\frac{{\partial} X}{{\partial} t}} + \frac{1}{2} {\frac{{\partial}^{2} X}{{\partial} W^{2}}} \quad ; \quad \sigma X = {\frac{{\partial} X}{{\partial} W}}.$ From the second equation, we note that $X \propto e^{\sigma W} \quad \text{ and therefore } \quad {\frac{{\partial}^{2} X}{{\partial} W^{2}}} = \sigma^{2} X.$ Substituting into the first equation, it follows that ${\frac{{\partial} X}{{\partial} t}} = \bigg(\mu - \frac{1}{2} \sigma^{2}\bigg) X.$ Recognizing an exponential as the solution class for $X(t)$ again, we arrive at the unique solution: $X(t) = X_{0} e^{\big(\mu - \frac{1}{2}\sigma^{2}\big) t + \sigma W(t)}.$

### Properties

Recall that $W(t) \sim \mathcal{N}(0, t)$.

It follows that, for constant $\mu$ and $\sigma$, $X(t)$ describes a Galton distribution, i.e., $X(t) \sim X_{0}\exp\left( \Big(\mu - \frac{1}{2}\sigma^{2}\Big) t + Z \sigma \sqrt{t} \right),$ for $Z \sim \mathcal{N}(0, 1)$.

It follows that $\mathbb{E}[X_{t}] = X_{0} e^{\mu t} \quad ; \quad {\rm Var}[X_{t}] = e^{2 \mu t} \bigg(e^{(\sigma^{2} t)} - 1\bigg).$

### vs Discrete-Time

Imagine investing in a security $X$, the valuation of which (e.g., relative to USD) grows by a ratio $r_{t} \sim \mathcal{N}(\tilde{\mu}, \tilde{\sigma}^{2})$ each period, for constants $\tilde{\mu}$ and $\tilde{\sigma}$.

Should we model the security according to geometric Brownian motion, which applies in the continuous-time limit, or is this inappropriate when changes happen in discrete time?

First, what is the statistical behavior of $X$ after $t$ periods when the process evolves discretely? For $X_{t} = X_{0} \prod_{s=0}^{t-1}r_{s} \quad ; \quad r_{s} \sim \mathcal{N}(\tilde{\mu}, \tilde{\sigma}^{2}) \quad \text{(independently)}$ we have $\mathbb{E}[X_{t}] = X_{0} \tilde{\mu}^{t} \quad ; \quad {\rm Var}[X_{t}] = \tilde{\mu}^{2t} \bigg(\Big(\frac{\tilde{\sigma}^{2}}{\tilde{\mu}^{2}} + 1\Big)^{t} - 1 \bigg)$

The statistics of this process agree with those of geometric Brownian motion when we identify $\mu = \log \tilde{\mu} \quad ; \quad \sigma^{2} = \log \left( \frac{\tilde{\sigma}^{2}}{\tilde{\mu}^{2}} + 1 \right).$